If `((1+i)/(1-i))^x = 1,` then

Options

A. `x=2n+1`, where n is any positive integer.
B. x=4n, where n is any positive integer
C. x=2n, where n is any positive integer
D. x=4n+1, where n is any positive integer.

Explanation

To solve the equation \(\left(\frac{1+i}{1-i}\right)^x = 1\), we will follow these steps:

### Step 1: Simplify the expression \(\frac{1+i}{1-i}\)

To simplify \(\frac{1+i}{1-i}\), we will multiply the numerator and the denominator by the conjugate of the denominator:

\[ \frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} \]

### Step 2: Calculate the numerator and denominator

**Numerator:**
\[ (1+i)(1+i) = 1^2 + 2 \cdot 1 \cdot i + i^2 = 1 + 2i - 1 = 2i \]

**Denominator:**
\[ (1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2 \]

### Step 3: Combine the results

Now we can combine the results from the numerator and denominator:

\[ \frac{1+i}{1-i} = \frac{2i}{2} = i \]

### Step 4: Substitute back into the equation

Now we substitute this back into the original equation:

\[ (i)^x = 1 \]

### Step 5: Analyze the equation \(i^x = 1\)

The complex number \(i\) can be expressed in exponential form as:

\[ i = e^{i\frac{\pi}{2}} \]

Thus, we can write:

\[ i^x = \left(e^{i\frac{\pi}{2}}\right)^x = e^{i\frac{\pi}{2}x} \]

For \(e^{i\frac{\pi}{2}x} = 1\), the exponent must be an integer multiple of \(2\pi\):

\[ \frac{\pi}{2}x = 2n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 6: Solve for \(x\)

Solving for \(x\):

\[ x = \frac{2n\pi}{\frac{\pi}{2}} = 4n \]

where \(n\) is any integer.

### Conclusion

Thus, \(x\) must be a multiple of \(4\):

\[ x = 4n \quad \text{where } n \in \mathbb{Z}^+ \]